Simplify and expand the following expression: $ \dfrac{5}{4q + 8}- \dfrac{4}{3q + 9}- \dfrac{3}{q^2 + 5q + 6} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{5}{4q + 8} = \dfrac{5}{4(q + 2)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{4}{3q + 9} = \dfrac{4}{3(q + 3)}$ We can factor the quadratic in the third term: $ \dfrac{3}{q^2 + 5q + 6} = \dfrac{3}{(q + 2)(q + 3)}$ Now we have: $ \dfrac{5}{4(q + 2)}- \dfrac{4}{3(q + 3)}- \dfrac{3}{(q + 2)(q + 3)} $ The least common multiple of the denominators is: $ 12(q + 2)(q + 3)$ In order to get the first term over $12(q + 2)(q + 3)$ , multiply by $\dfrac{3(q + 3)}{3(q + 3)}$ $ \dfrac{5}{4(q + 2)} \times \dfrac{3(q + 3)}{3(q + 3)} = \dfrac{15(q + 3)}{12(q + 2)(q + 3)} $ In order to get the second term over $12(q + 2)(q + 3)$ , multiply by $\dfrac{4(q + 2)}{4(q + 2)}$ $ \dfrac{4}{3(q + 3)} \times \dfrac{4(q + 2)}{4(q + 2)} = \dfrac{16(q + 2)}{12(q + 2)(q + 3)} $ In order to get the third term over $12(q + 2)(q + 3)$ , multiply by $\dfrac{12}{12}$ $ \dfrac{3}{(q + 2)(q + 3)} \times \dfrac{12}{12} = \dfrac{36}{12(q + 2)(q + 3)} $ Now we have: $ \dfrac{15(q + 3)}{12(q + 2)(q + 3)} - \dfrac{16(q + 2)}{12(q + 2)(q + 3)} - \dfrac{36}{12(q + 2)(q + 3)} $ $ = \dfrac{ 15(q + 3) - 16(q + 2) - 36} {12(q + 2)(q + 3)} $ Expand: $ = \dfrac{15q + 45 - 16q - 32 - 36}{12q^2 + 60q + 72} $ $ = \dfrac{-q - 23}{12q^2 + 60q + 72}$